Question

An object is thrown vertically upwards to a height of $10\mathrm{m}$. Calculate its time taken by the object to reach the highest point.

Answer 1

Step 1: Given data:

Distance travelled, $\mathrm{s}=10\mathrm{m}$,

Final velocity, $\mathrm{v}=0\mathrm{m}/\mathrm{s}$

$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

We have to calculate the velocity with which it was thrown upward and the time taken by the object to reach the highest point.

Step 2: Formula to be used.

According to the third equation of motion,

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gs}$

According to the first equation of motion,

$\mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{g}}$

Here, $\mathrm{u}$ is initial velocity, $\mathrm{g}$ is the acceleration due to gravity, and $\mathrm{v}$ is final velocity and $\mathrm{t}$ is time.

Step 3: Calculate the velocity with which it was thrown upward.

So,

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gs}$

${\left(0\right)}^2-{\mathrm{u}}^2=2\times 9.8\times 10$

${\mathrm{u}}^2=196$

$\mathrm{u}=\sqrt{196}$

$\mathrm{u}=14\mathrm{m}/\mathrm{s}$

So, the velocity with which the object is thrown upwards is $14\mathrm{m}/\mathrm{s}$.

Step 4: Calculate the time taken by the object to reach the highest point.

The velocity with which the object is thrown upwards is $14\mathrm{m}/\mathrm{s}$. So,

$\mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{g}}$

$\mathrm{t}=1.43\sec$

Hence, the time taken by the object to reach the highest point is $1.43\sec$.